OFFSET
1,3
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 1. Also, for any integer n > 4, p(k)*q(n-k) - 1 is prime for some 0 < k < n/2.
(ii) If n > 9, then prime(k)*p(n-k) + 1 is prime for some 0 < k < n. If n > 2, then prime(k)*q(n-k) - 1 is prime for some 0 < k < n, and also prime(k)*q(n-k) + 1 is prime for some 0 < k < n.
(iii) If n > 11, then prime(k) + p(n-k) is prime for some 0 < k < n. If n > 4, then prime(k) + q(n-k) is prime for some 0 < k < n, and also prime(k)^2 + q(n-k)^2 is prime for some 0 < k < n.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Z.-W. Sun, On a^n+ bn modulo m, arXiv preprint arXiv:1312.1166 [math.NT], 2013-2014.
EXAMPLE
a(5) = 1 since 5 = 1 + 4 with p(1)^2 + q(4)^2 = 1^2 + 2^2 = 5 prime.
a(6) = 1 since 6 = 3 + 3 with p(3)^2 + q(3)^2 = 3^2 + 2^2 = 13 prime.
a(19) = 1 since 19 = 3 + 16 with p(3)^2 + q(16)^2 = 3^2 + 32^2 = 1033 prime.
a(28) = 1 since 28 = 3 + 25 with p(3)^2 + q(25)^2 = 3^2 + 142^2 = 20173 prime.
a(29) = 1 since 29 = 6 + 23 with p(6)^2 + q(23)^2 = 11^2 + 104^2 = 10937 prime.
a(38) = 1 since 38 = 1 + 37 with p(1)^2 + q(37)^2 = 1^2 + 760^2 = 577601 prime.
a(75) = 1 since 75 = 13 + 62 with p(13)^2 + q(62)^2 = 101^2 + 13394^2 = 179409437 prime.
a(160) = 1 since 160 = 48 + 112 with p(48)^2 + q(112)^2 = 147273^2 + 1177438^2 = 1408049580373 prime.
a(210) = 1 since 210 = 71 + 139 with p(71)^2 + q(139)^2 = 4697205^2 + 8953856^2 = 102235272080761 prime.
MATHEMATICA
a[n_]:=Sum[If[PrimeQ[PartitionsP[k]^2+PartitionsQ[n-k]^2], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 07 2013
STATUS
approved